The derivative of a function defined by an integral

Let $f$ be a continuous function on $\mathbb R$. Let $\displaystyle g(t)=\int_0^t f(x)\;dx$, the fundamental theorem of calculus says that $g'(t)=f(t)$. Is this true for any $t$ or only for $t>0$ ? I mean if $t<0$ should we write $g'(t)=-f(t)$ or it will be the same $g'(t)=f(t)$ even if $t<0$ ?

Solution 1:

The easiest way to figure this question out is to prove that it holds for any $t$. So let $t \in \mathbb{R}$, and $h \in \mathbb{R}^*$. One has \begin{align*} \left|\frac{g(t+h)-g(t)}{h} - f(t)\right| &= \left|\frac{1}{h} \int_t^{t+h} f(x) dx - f(t)\right|\\ & = \left|\frac{1}{h} \int_t^{t+h} (f(x)-f(t))dx \right|\\ & \leq \frac{1}{|h|} \left|\int_t^{t+h} |f(x)-f(t)| dx \right|\\ & \leq \sup_{x \in [t,t+h]} |f(x)-f(t)| \end{align*}

Because $f$ is continuous at the point $x=t$, then the RHS tends to $0$ as $h$ tends to $0$, so you get that $$\lim_{h \rightarrow 0} \frac{g(t+h)-g(t)}{h} = f(t), \quad \quad \text{i.e.} \quad \boxed{g'(t) = f(t)}$$

You can see that this proof works for $t < 0$ as well as for $t \geq 0$.